By Sunil Bhardwaj

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TLC

Consider a diatomic molecule AB. Two atoms A and B are connected by an elastic spring. A is fixed at one end and B is kept hanging at their normal equilibrium distance \(r\). When the spring is stretched by an external force \(f\) and then released, the equilibrium distance is disturbed, and its length increases. The opposing force restores back the spring to its original equilibrium distance. Thus, the atom B performs simple harmonic motion along the axis of the bond.

When bond atoms A and B are vibrating and oscillating simultaneously then the vibrational frequency \((\mu )\) is equal to oscillating frequency \((\omega )\), is given by the relation, $$\omega = v = \frac { 1 }{ 2\pi } \sqrt { \frac { K }{ \mu } } $$

On the basis of quantum mechanical treatment and application of Schrondinger wave equation ,it is found that the energy associated with vibration of a molecule is quantised and the vibrational energy of a harmonic oscillator is given by, $${ E }_{ v } = \left( v + \frac { 1 }{ 2 } \right) h\omega$$ where \(v\) = vibrational quantum number having values 0,1,2,.....etc. \(h\) = Plancks constant.

For lowest energy level, \(v = 0 \) $${ E }_{ v } = \frac { 1 }{ 2 } h\omega$$ For different vibrational quantum number, corresponding energy levels will have value as

TLC

For \(v = 1\), $$ { E }_{ 1 } = \left( 1 + \frac { 1 }{ 2 } \right) h\omega \qquad { E }_{ 1 } = \frac { 3 }{ 2 } h\omega$$ For \(v = 2\), $$ { E }_{ 2 } = \left( 2 + \frac { 1 }{ 2 } \right) h\omega \qquad { E }_{ 2 } = \frac { 5 }{ 2 } h\omega$$ For \(v = 3\), $$ { E }_{ 3 } = \left( 3 + \frac { 1 }{ 2 } \right) h\omega \qquad { E }_{ 3 } = \frac { 7 }{ 2 } h\omega $$ For \(v = 4\), $$ { E }_{ 4 } = \left( 4 + \frac { 1 }{ 2 } \right) h\omega \qquad { E }_{ 4 } = \frac { 9 }{ 2 } h\omega$$ and so on.

The difference between two successive vibrational energy levels can be obtained as follows.
For the transition \(v\) to \(v'\) we get, $$ \Delta { E }_{ vib } = { E }_{ v' } - { E }_{ v } $$ $$= \left( v' + \frac { 1 }{ 2 } \right) h\omega - \left( v + \frac { 1 }{ 2 } \right) h\omega $$ $$ = \left( v' + \frac { 1 }{ 2 } - v - \frac { 1 }{ 2 } \right) h\omega $$ $$ = \left( v' - v \right) h\omega $$ for the transition \(v = 1\) to \(v = 2\) $$\Delta { E }_{ vib } = \left( v' - v \right) h\omega $$ $$ = \left( 2 - 1 \right) h\omega = h\omega $$ $$ = \frac { h }{ 2\pi } \sqrt { \frac { K }{ \mu } } $$ It can be understood from this relation that the vibrational energy levels are also equally spaced having a constant energy difference of \(h\omega\) between any two successive energy levels and the spacing between the levels depends on the force constant \((K)\) and the reduced mass \((\mu )\).

Calculate zero point energy and force constant of a molecule whose reduced mass is \(1.2 \times { 10 }^{ -27 }\) kg. The wave number of origin of the band in IR spectrum is \(37 \times { 10 }^{ 4 } { m }^{ -1 }\).

Solution: We have reduced mass \(\mu = 1.2 \times { 10 }^{ -27 } \) kg
wave number \(\overline { \upsilon } = 37 \times { 10 }^{ 4 } { m }^{ -1 }\)
Force constant K can be calculated from equation, $$ \overline { \upsilon } = \frac { 1 }{ 2\pi c } \sqrt { \frac { K }{ \mu } } $$ $$ \left( 37 \times { 10 }^{ 4 } { m }^{ -1 } \right) =\frac { 1 }{ 2 \times 3.14 \times \left( 3 \times { 10 }^{ 8 } m/s \right) } \sqrt { \frac { K }{ 1.2 \times { 10 }^{ -27 } kg } } $$ $$ \left( 37 \times { 10 }^{ 4 } { m }^{ -1 } \right) = \frac { 1 }{ 18.84 \times { 10 }^{ 8 } } \sqrt { \frac { K }{ 1.2 \times { 10 }^{ -27 } kg } } $$ $$ \sqrt { \frac { K }{ 1.2 \times { 10 }^{ -27 } kg } } = \left( 37 \times { 10 }^{ 4 } { m }^{ -1 } \right) \times \left( 18.84 \times { 10 }^{ 8 } \right) $$ $$ \sqrt { \frac { K }{ 1.2 \times { 10 }^{ -27 } kg } } = 6.971 \times { 10 }^{ 14 } $$ $$ \frac { K }{ 1.2 \times { 10 }^{ -27 } kg } = { \left( 6.971 \times { 10 }^{ 14 } \right) }^{ 2 } = 4.859 \times { 10 }^{ 29 }$$ $$ K = \left( 4.859 \times { 10 }^{ 29 } \right) \times \left( 1.2 \times { 10 }^{ -27 } \right) = 583.105 N{ m }^{ -1 } $$ Zero Point Energy is the vibrational energy at \(v = 0\) $$\therefore { E }_{ 0 } = \frac { 1 }{ 2 } h\omega = \frac { 1 }{ 2 } h\upsilon = \frac { 1 }{ 2 } h\overline { \upsilon } c $$ $$ \qquad = \frac { 1 }{ 2 } \left( 6.626 \times { 10 }^{ -34 } Js \right) \left( 37 \times { 10 }^{ 4 } { m }^{ -1 } \right) \left( 3 \times { 10 }^{ 8 } m/s \right) $$ $$ \qquad = \frac { 1 }{ 2 } \left( 7.355 \times { 10 }^{ -20 } \right) $$ $$ = 3.677 \times { 10 }^{ -20 } J $$

The vibrational frequency of HCl is \(2.988 \times { 10 }^{ 5 } { m }^{ -1 }\). Calculate the zero point energy of the molecule.

Solution: We have wave number \(\overline { \upsilon } = 2.988 \times { 10 }^{ 5 } { m }^{ -1 }\)
Zero Point Energy is the vibrational energy at \(v = 0\) $$ \therefore { E }_{ 0 } = \frac { 1 }{ 2 } h\omega = \frac { 1 }{ 2 } h\upsilon = \frac { 1 }{ 2 } h\overline { \upsilon } c $$ $$ \qquad = \frac { 1 }{ 2 } \left( 6.626 \times { 10 }^{ -34 } Js \right) \left( 2.988 \times { 10 }^{ 5 } { m }^{ -1 } \right) \left( 3 \times { 10 }^{ 8 } m/s \right) $$ $$ \qquad = \frac { 1 }{ 2 } \left( 5.94 \times { 10 }^{ -20 } \right) $$ $$ \qquad = 2.97 \times { 10 }^{ -20 } J $$

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